A239740 a(n) = gcd(Sum_{k=1...n} F(k), Product{j=1...n} F(j)), where F(k) is the k-th Fibonacci number.

1, 1, 2, 1, 6, 20, 3, 18, 8, 143, 8, 8, 21, 986, 84, 63, 220, 6764, 55, 770, 144, 46367, 144, 432, 377, 317810, 16588, 377, 43428, 2178308, 987, 53298, 2584, 14930351, 2584, 18088, 6765, 102334154, 784740, 20295, 2054476, 701408732, 17711, 1664834, 46368

Offset

1,3

Comments

The Fibonacci numbers in the sequence are 1, 2, 3, 8, 21, 55, 144, 377, 987, ... and a majority are elements of A001906 (F(2*n)= bisection of Fibonacci sequence).

We find consecutive values such that (1, 2), (2, 3), (20, 21), (986, 987), (6764, 6765), (46367, 46368), (317810, 317811), (14930351, 14930352), ...

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Example

The first 8 Fibonacci numbers are 1,1,2,3,5,8,13,21 and 1+1+2+3+5+8+13+21 = 54. So a(8) = gcd(54, 1*1*2*3*5*8*13*21) = 18.

Maple

with(combinat, fibonacci):seq(gcd(add(fibonacci(i), i=1..n), mul(fibonacci(j), j=1..n)), n=1..60);

Mathematica

nn=60; With[{prs=Fibonacci[Range[nn]]}, Table[GCD[Total[Take[prs, n]], Times@@Take[ prs, n]], {n, nn}]]

Prog

(Haskell)
a239740 n = gcd (sum fs) (product fs)
            where fs = take n $ tail a000045_list
-- Reinhard Zumkeller, Mar 27 2014

Crossrefs

Cf. A000045, A001906.

Sequence in context: A365109 A025271 A153804 * A371986 A268371 A318918

Adjacent sequences: A239737 A239738 A239739 * A239741 A239742 A239743

Keyword

nonn

Author

Michel Lagneau, Mar 26 2014

Status

approved